What is the limit as z approaches-3 of (z^2-9)/(z+3) ?

The limit as z approaches-3 of (z^2-9)/(z+3) is -6

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limit as z approaches-3 of (z^2-9)/(z+3)

z \to - 3 lim (\frac{z ^{2} - 9}{z + 3})

- 6

Solution steps

z \to - 3 lim (\frac{z ^{2} - 9}{z + 3})

Simplify

\frac{z ^{2} - 9}{z + 3} : z - 3

= z \to - 3 lim (z - 3)

Plug in the value

z = - 3

= - 3 - 3

Simplify

= - 6

\int sec^{6} (7 x) d x

y^{^{'}} = 0.0012 y (1900 - y)

\frac{\partial}{\partial x} (x^{3} + y^{3} + z^{3} - 3 x yz)

\frac{d y}{d x}, y^{\frac{2}{3}} - x^{2} + x y = 1

\frac{\partial}{\partial x} (5 x sin (3 x^{2} y))

What is the limit as z approaches-3 of (z^2-9)/(z+3) ?
The limit as z approaches-3 of (z^2-9)/(z+3) is -6